8
$\begingroup$

Is there a group $G$ and subgroups $H_1, \dots, H_n \leq G$ for some $n \in \mathbb{N}$, such that $[G : H_i] = \infty$ for each $1 \leq i \leq n$, and $$G = \bigcup_{i=1}^n H_i \ \ ?$$

Improve this question
$\endgroup$

1 Answer 1

Reset to default
13
$\begingroup$

It may well be a lot easier than this, but it follows from the answer to the weaker question Can a group be a finite union of (left) cosets of infinite-index subgroups? that it's not possible.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Looks good to me. $\endgroup$
    – Pete L. Clark
    Commented Sep 7, 2015 at 16:51
  • 5
    $\begingroup$ More precisely, the claim follows from Lemma 4.2 of Newman's paper in math.uga.edu/~pete/Neumann54.pdf , which has a half-page proof (basically by induction on the number of distinct subgroups, the right-cosets of which one is trying to use to cover the whole group). $\endgroup$
    – Terry Tao
    Commented Sep 7, 2015 at 21:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .